- #1
TheSodesa
- 224
- 7
Homework Statement
Show that the largest possible change in the kinetic energy , ##\Delta E_{kin}##, of a particle of mass ##m## running into another particle of mass ##M## at rest in the lab coordinate system is
[tex]\Delta E_{kin} = \frac{-4AE_{kin}}{(1+A)^{2}}[/tex], where ##A = \frac{M}{m}##.
Calculate the limit [tex]\lim_{M \to m} \Delta E_{kin}[/tex] and plot ##\Delta E_{kin}(A)##.
Homework Equations
Conservation of energy.
Conservation of momentum.
Rutherford's experiment?
The Attempt at a Solution
I'm going to assume, that both energy and momentum are conserved.
Energy:
[tex]
E_{mi} + \stackrel{0}{E_{Mi}} = E_{mf} + E_{Mf}\\ \iff \\ |\Delta E_{kin}| = |E_{mf} - E_{mi}| = |-E_{Mf}| = \frac{1}{2}Mv^{2}
[/tex]
Momentum:
\begin{align*}
x) &&mu_{m} &= mv_{m}\cos{\theta} + Mv_{M}\cos{\varphi}\\
y) &&0 &= mv_{m}\sin{\theta} + Mv_{M}\sin{\varphi}
\end{align*}
Adding these equations together and solving for ##v_{M}## yields
\begin{align*}
v_{M}
&= \frac{mu_{m} - mv_{m}(\sin{\theta} + \cos{\theta})}{M(\sin{\varphi} + \cos{\varphi})} &| \sin{x} + \cos{x} = \sqrt{2}\sin{(x + \frac{\pi}{4})}\\
&= \frac{m(u_{m} - \sqrt{2}v_{m}\sin{(\theta + \frac{\pi}{4})})}{\sqrt{2}M\sin{(\varphi + \frac{\pi}{4})}}
\end{align*}
Plugging this into the energy expression:
\begin{align*}
\Delta E_{kin}
&= \frac{1}{2}M \left( \frac{m(u_{m} - \sqrt{2}v_{m}\sin{(\theta + \frac{\pi}{4})})}{\sqrt{2}M\sin{(\varphi + \frac{\pi}{4})}} \right)^{2} \\
&= \frac{m^{2}(u_{m} - \sqrt{2}v_{m}\sin{(\theta + \frac{\pi}{4})})^{2}}{4M\sin^2 (\varphi + \frac{\pi}{4})} \\
\end{align*}
Now the numerator is at it's largest, when (this bit is irrelevant)
[tex]
sin(\theta + \frac{\pi}{4}) = 0 \\
\iff\\
\theta + \frac{\pi}{4}rad = 0^{\circ} + n \times 180^{\circ}\\
\iff\\
\theta = 180^{\circ} - \frac{\pi}{4}rad = 180^{\circ} - 45^{\circ} = 135^{\circ}
[/tex]
Then, if we differentiate with respect to ##\varphi##,
\begin{align*}
\frac{d \Delta E_{kin}}{d \varphi}
&= \frac{d}{d \varphi} \frac{m^{2}(u_{m})^{2}}{4M\sin^{2} (\varphi + \frac{\pi}{4})} &&| \text{ Let }\frac{m^{2}(u_{m})^{2}}{4M} = C \\
&= \frac{d}{d \varphi} C \sin^{-2}(\varphi + \frac{\pi}{4})\\
&= -2 C \cos(\varphi + \frac{\pi}{4})\sin^{-3}(\varphi + \frac{\pi}{4})\\
&= 0
\end{align*}
when
[tex]
cos(\varphi + \frac{\pi}{4}) = 0\\
\iff\\
\varphi + \frac{\pi}{4} = \frac{\pi}{2}\\
\iff\\
\varphi = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} = 45^{\circ}
[/tex]
Alright, now we have the angles (which I'm not convinced by). Now what? Plugging them back into the energy expression isn't going to help me simplify it.
I feel like I've gone down a wrong path. Help, please?